java 谐波序列递归
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Harmonic sequence recursion
提问by vaindil
I'm really getting the hang of recursion (or so I think), but this problem is tripping me up. I'm trying to return 1 + 1/2 + 1/3 + ... + 1/n, but no matter what I try the method returns 1.0. I cannot for the life of me figure out what's wrong.
我真的掌握了递归的窍门（或者我认为是这样），但是这个问题让我很困惑。我试图返回 1 + 1/2 + 1/3 + ... + 1/n，但无论我尝试什么，该方法都会返回 1.0。我一生都无法弄清楚出了什么问题。
public static double harmonic(int n) {
if(n == 1) {
return 1;
} else {
return (1 / n) + (1 / harmonic(n - 1));
}
}
采纳答案by Brian
Well, for one, you don't want to return (1 / n) + (1 / harmonic(n - 1)), but also you need to use doublearithmetic:
嗯，一方面，你不想 return (1 / n) + (1 / harmonic(n - 1))，但你也需要使用double算术：
public static double harmonic(int n) {
if(n == 1) {
return 1.0;
} else {
return (1.0 / n) + harmonic(n - 1);
}
}
If you left it as 1 / harmonicyou'd return another function entirely:
如果你离开它，1 / harmonic你会完全返回另一个函数：
(1 / n) + 1 / ( 1 / (n - 1) + 1 / ( 1 / (n - 2) + 1 / (...) ) )
(1 / n) + 1 / ( 1 / (n - 1) + 1 / ( 1 / (n - 2) + 1 / (...) ) )
That is a very confusing function to figure out, btw, but I think (with my 3rd time editing it) I got it right this time.
顺便说一句，这是一个非常令人困惑的功能，但我认为（我第三次编辑它）这次我做对了。
回答by Claudiu
You want to use floating point division:
你想使用浮点除法：
public static double harmonic(int n) {
if(n == 1.0) {
return 1.0;
} else {
return (1.0 / n) + (1.0 / harmonic(n - 1.0));
}
}
That is: 1/2is 0; 1/2.0is 0.5.
即：1/2是0；1/2.0是0.5。
回答by Rohit Jain
Thats because integer division gives integer result.
那是因为整数除法给出整数结果。
So, 1/2 == 0
所以， 1/2 == 0
You can use rather use floating-pointdivision like this: -
您可以floating-point像这样使用除法： -
if(n == 1.0) {
return 1.0;
} else {
return (1.0 / n) + harmonic(n - 1); // Should be `harmonic(n - 1)`
}
回答by paddy
You need to use doubles. Right now, you're doing 1 / n, both of which are integers. Change it to:
你需要使用双打。现在，您正在执行1 / n，这两个都是整数。将其更改为：
return (1.0 / n) + (1.0 / harmonic(n - 1));
回答by FThompson
Use doubles in your division calculations. Currently, everything is cast to ints, losing any floating-point precision you would normally expect.
在除法计算中使用双精度数。目前，一切都被转换为整数，失去了您通常期望的任何浮点精度。
public static double harmonic(int n) {
if (n == 1) {
return 1;
} else {
return (1.0 / n) + (1.0 / harmonic(n - 1));
}
}
回答by Seid.M
the recursion part should not include 1/harmonic(n-1)it should be
递归部分不应该包括1/harmonic(n-1)它应该是
public static double harmonic(int n)
{
double harm = 0.0;
if (n == 1) {
return 1.0;
}
else {
harm = harm + (1.0 / n) + harmonic(n - 1);
}
return harm;
}
回答by Hrishikesh Mishra
/**
* Created by hrishikesh.mishra on 04/01/16.
*
* Describe a recursive algorithm
* for computing the nth Harmonic number,
* defined as Hn = ∑ n k=1 1/k.
*
*/
public class HarmonicNumber {
public static void main(String[] args) {
System.out.println("Sum up to 1: " + sum(1));
System.out.println("Sum up to 2: " + sum(2));
System.out.println("Sum up to 3: " + sum(3));
System.out.println("Sum up to 4: " + sum(4));
}
/**
* Summation with recursive method.
* @param n
* @return
*/
public static double sum(int n){
if(n <= 1)
return 1;
else
return ((double) 1/n) + sum(n - 1);
}
}
回答by Mahmoud
public static double harmonic(int n)
{
if (n==1)
return 1/n;
else return 1/n + harmonic(n-1);
}
